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\begin{document}
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Consider the following coordinate systems:

\begin{eqnarray}
x &=& \sin(\theta)\cos(\phi) \nonumber \\
y &=& \sin(\theta) \sin(\phi) \nonumber \\
z &=& \cos(\theta) \nonumber
\end{eqnarray}

\begin{eqnarray}
x &=& \sin(\eta)\cos(\xi) \nonumber \\
y &=& \cos(\eta) \nonumber \\
z &=& \sin(\eta) \sin(\xi) \nonumber
\end{eqnarray}

We want to use as coordinate patch near $x=1$ on the unit sphere the
coordinates
$(\phi, \xi)$. To find the surface element we need only to find $\theta$
as a
function of these two coordinates.
Since the surface element is:

\[
ds^2 = d\theta^2 + \sin^2(\theta) d\phi^2
\]

Now,
\begin{eqnarray}
\tan(\xi) &=& z/x = 1/(\tan{\theta} \cos(\phi))  \nonumber \\
\end{eqnarray}

Using $(1+\tan^2(\alpha)) = 1/\cos^2(\alpha)$ (for any $\alpha$) we get

\[
\tan(\theta) = (1+\tan^2(\phi))^{1/2}/\tan(\xi)
\]

and defining $a=\tan(\xi)$;  $b=\tan(\phi)$ we get

\[
d\theta = \frac{\sqrt{1+b^2}}{1+a^2+b^2}[abd\phi - (1+a^2) d\xi]
\]

Since
\begin{eqnarray}
\sin^2(\theta) &=& \frac{\tan^2(\theta)}{1+\tan^2(\theta)} \\
               &=& \frac{(1+b^2)}{1+a^2+b^2}
\end{eqnarray}

and the surface element becomes:

\[
ds^2 = \frac{(1+a^2)(1+b^2)}{(1+a^2+b^2)^2}[(1+a^2) d\xi^2 + (1+b^2)
d\phi^2 - 2 ab d\xi d\phi]
\]

Here everything is nice and regular and symmetric.

The Jacovian. We only need the Jacovian at one side, all the others follow from 
symmetry. We do it at the boundary between the patch $x=1$ and the patch $y=1$.
For the first we have (as above) :

\begin{eqnarray}
x &=& \sin(\theta)\cos(\phi) \nonumber \\
y &=& \sin(\theta) \sin(\phi) \nonumber \\
z &=& \cos(\theta) \nonumber
\end{eqnarray}

\begin{eqnarray}
x &=& \sin(\eta)\cos(\xi) \nonumber \\
y &=& \cos(\eta) \nonumber \\
z &=& \sin(\eta) \sin(\xi) \nonumber
\end{eqnarray}

For the second:

\begin{eqnarray}
x &=& -\sin(\tilde{\theta})\sin(\tilde{\phi}) \nonumber \\
y &=& \sin(\tilde{\theta})\cos(\tilde{\phi}) \nonumber \\
z &=& \cos(\tilde{\theta}) \nonumber
\end{eqnarray}

\begin{eqnarray}
y &=& \sin(\eta')\cos(\xi') \nonumber \\
x &=& \cos(\eta') \nonumber \\
z &=& \sin(\eta')\sin(\xi') \nonumber
\end{eqnarray}

Therefore:

\begin{eqnarray}
\frac{x}{y}&=& \tan(\tilde{\phi}) = -\frac{1}{\tan(\phi)}  \nonumber \\
\frac{z}{y}&=& \tan(\xi') = \frac{1}{\tan(\theta)\sin(\phi)} = \frac{\tan(\xi)}{\tan(\phi)}= \frac{a}{b}  \nonumber 
\end{eqnarray}

Thus,
\begin{eqnarray}
d\tilde{\phi} &=& d\phi \nonumber \\
d\xi'&=& \frac1{a^2+b^2}[b(1+a^2) d\xi - a(1+b^2)d\phi] \nonumber 
\end{eqnarray}

At the boundary $\phi=\frac{\pi}{4}$ $\tilde{\phi}= -\frac{\pi}{4}$, and $b=1$,
so $d\xi'= d\xi - \frac{2a}{1+a^2}d\phi$

The inverse transformation is $d\xi = d\xi' + \frac{2a}{1+a^2}d\tilde{\phi}$


If we take as variables $a$, and $b$ the metric becomes:

\[
ds^2 = \frac{1}{1+a^2+b^2}[(1+b^2)da^2 + (1+a^2)db^2 - 2ab da db]
\]


The inverse metric is:

\begin{equation}
g^{AB} = D^2 \left(\begin{array}{cc}
            1+a^2 & ab \\
            ab   & 1+b^2
        \end{array}
      \right)
\end{equation}
and
\[
det(g_{AB}) = 1/ D^6
\]

Where $D:= \sqrt{1+a^2+b^2}$.
In the computation of the Laplacian we need $\sqrt{g}g^{AB}$ which is
given by
    
\begin{equation}
\sqrt{g}g^{AB} = \frac{1}{D} \left(\begin{array}{cc}
            1+a^2 & ab \\
            ab   & 1+b^2
        \end{array}
      \right)
\end{equation}

For numerical integration it is convenient to introduce the co-vector:

\[
W_A := \partial_A \phi
\]

and the vector 

\[
\tilde{W}^A := \sqrt{g}g^{AB}W_B = \frac{1}{D}((1+a^2)W_a + abW_b, (1+b^2)W_b + abW_a)
\]

Then 

\[
\Delta \phi = D^3(\partial_a(\tilde{W}^a) + \partial_b(\tilde{W}^b)) 
\]

\section{New version}

Set $a := x/|z|$, $b := y/|z|$, $z := \sqrt{1-x^2-y^2} := 1/\sqrt{1 + a^2 + b^2}:= \frac{1}{D}$

Thus, 
$x = \frac{a}{\sqrt{1 + a^2 + b^2}}$, 
$dx = \frac{da}{D} - \frac{a^2 da}{D^3} - \frac{ab db}{D^3} = \frac{(1+b^2)da - ab db}{D^3}$

similarly
$dy = \frac{(1+a^2)db - ab da}{D^3}$, 
and
$dz = -\frac{ada + bdb}{D^3}$

Thus,

\begin{eqnarray}
ds^2 &=& dx^2 + dy^2 + dz^2 \nonumber \\
     &=&  \frac{1}{D^6}[[(1+b^2)^2 + a^2b^2 + a^2]da^2 + [(1+a^2)^2 + a^2b^2 + b^2]db^2\nonumber \\
     & & - 2ab[(1+b^2) + (1+a^2) - 1]da db\nonumber \\
     &=& \frac{1}{D^4}[(1+b^2)^2da^2 + (1+a^2)^2db^2 - 2abda db] 
\end{eqnarray}


The boundary matrix at boundary $a=\pm 1$ is $A^A n_A$ where 
$n_A = (\frac{nx}{D\sqrt{1+a^2}}, 0)= (\frac{n_x}{D\sqrt{2}}, 0)$ where $n_x = \pm 1$ acordingly to
which boundary we are at.

\begin{equation}
A^A n_A = \frac{Dn_x}{\sqrt{1+a^2}}\left(\begin{array}{ccc}
                                        0 & 1+a^2 & ab \\
                                        1 & 0 & \\
                                        0 & 0 & 
                                   \end{array}
                                   \right)
\end{equation}


The eigenvectors are $U^{\pm} = (\sqrt{1+a^2}, \pm n_x, 0)$ and $U^0 = (0,ab , -(1+a^2))$.

The co-basis is:
$\theta_{\pm} = \frac{1}{2}(\frac{1}{\sqrt{1+a^2}}, \pm nx, \frac{\pm abn_x}{1+a^2})$

The eigenvectors are: $\lambda_{\pm} = \pm D$, $\lambda_0 = 0$.
\end{document}